%I #43 Oct 08 2023 04:53:11
%S 4,28,70,130,208,304,418,550,700,868,1054,1258,1480,1720,1978,2254,
%T 2548,2860,3190,3538,3904,4288,4690,5110,5548,6004,6478,6970,7480,
%U 8008,8554,9118,9700,10300,10918,11554,12208,12880,13570,14278,15004
%N a(n) = (3*n+1)*(3*n+4).
%D L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 38.
%H Vincenzo Librandi, <a href="/A085001/b085001.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F Sum_{k=0..n} 3/a(k) = 3*(n+1)/(3n+4). [Corrected by _Gary Detlefs_, Mar 14 2018]
%F Sum_{k>=0} 3/a(k) = 1.
%F From _Gary W. Adamson_, Jan 03 2007: (Start)
%F Sum_{k>=0} 1/a(k) = 1/3.
%F Sum_{k=0..n} 1/a(k) = (n+1)/(3n+4) [Jolley]. (End) [Corrected by _Gary Detlefs_, Mar 14 2018]
%F G.f.: 2*(2+8*x-x^2)/(1-x)^3. - _R. J. Mathar_, Sep 17 2008
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Jul 08 2012
%F Sum_{n>=0} (-1)^n/a(n) = 2*Pi/(9*sqrt(3)) + 2*log(2)/9 - 1/3. - _Amiram Eldar_, Oct 08 2023
%t CoefficientList[Series[2*(2+8x-x^2)/(1-x)^3,{x,0,50}],x] (* _Vincenzo Librandi_, Jul 08 2012 *)
%t Table[(3n+1)(3n+4),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{4,28,70},50] (* _Harvey P. Dale_, Apr 07 2019 *)
%o (Magma) [(3*n+1)*(3*n+4): n in [0..50]]; // _Vincenzo Librandi_, Jul 08 2012
%o (PARI) a(n)=(3*n+1)*(3*n+4) \\ _Charles R Greathouse IV_, Jun 17 2017
%K nonn,easy
%O 0,1
%A _Gary W. Adamson_, Jun 17 2003
%E Edited by _Don Reble_, Nov 13 2005